Chapter 3
Quadratic Models
Ch 3 Quadratic Equation
A quadratic equation involves the square of
the variable. It has the form
y = ax 2 + bx + c where a, b and c are
constants
If a = 0 , there is no x-squared term, so the
equation is not quadratic
Graph of the quadratic equation y = 2x2 – 5
To solve the equation 2x2 – 5 = 7
We first solve for x2 to get
2x2 = 12
x2 = 6
x=+
-
x
-3
-2
-1
0
1
2
3
y
13
3
-3
-5
-3
3
13
6 = + 2.45 and – 2.45
7
5
3
1
-3
-2
-1
0
1
2
+3
Extraction of roots (Ex, 1 Pg 145)
The formula
h
h= 20 - 16t2
when t = 0.5
= 20 – 16(0.5) 2
= 20 – 16(0.25)
= 20 – 4
= 16ft
a (16, 0.5)
20
When h = 0 the equation to obtain
10
0 = 20 - 16t 2
16t 2 = 20
b
0.5
Time
1
t2
1.5
t
= 20/16 = 1.25
t = + 1.25 = + 1. 118sec
-
-
Solving Formulas
Volume of Cone
V=1
 r2
h
h
3
3V=  r2 h ( Divide both sides by h )
and find square root
r= +
-
r
3V
h
r
Volume of Cylinder V=  r2 h
V = r2
(Dividing both sides
h
by  h )
r= +
V
h
h
Compound Interest Formula
• A = P(1 + r) n
Where A = amount, P = Principal, R = rate of
Interest, n = No.of years
More Extraction of Roots
Equations of the form
a( x – p) 2 = q
Can also be solved by extraction of roots
after isolating the squared expression
( x – p) 2
Pythagorian Formula for Right
angled triangle
A
Hypotenuse
Height
90 degree
B
C
Base
In a right triangle
(Hypotenuse) 2 = (Base) 2 +(Height) 2
What size of a square can be inscribed in a circle of radius 8 inches ?
16 inches
8 in
s
8in
s
s represent the length of a side of the square
s 2 + s 2 = 16 2
2s = 256
s 2 = 128
s = 128 = 11.3 inches
Ch 3.2 Some examples of Quadratic Models
Height of a Baseball (Pg 156)
H = -16t 2 + 64t + 4
Evaluate the formula to complete the table of values for the
height of the baseball
70
Highest point
t
0
1
2
3
4
h
4
52
68
52
4
60
50
3) After ½ second base ball height h = -16(1/2) 2 +
64(1/2) + 4 = 32 ft
4) 3.5 second height will be 32 ft
40
30
5) When the base ball height is 64 ft the time will
be 1.5 sec and 2.5 sec
20
10
0
1
2
3
4
6) When 20 ft the time is 0.25 and 3.75 sec
7) The ball caught = 4 sec
Example 4 ( Page 156 )
• Using Graphing Calculator
• H = - 16 x2 + 64x + 4
Press Y key
table
TblStart = 0 and increment 1
Press graph
Press 2nd ,
3.3 Solving Quadratic Equations by
Factoring
Zero Factor Principle
The product of two factors equals zero if and only if one or
both of the factors equals zero.
In symbols ab = 0 if and only if a = o or b = 0
Example (x – 6) (x + 2) = 0
x – 6 = 0 or x + 2 = 0
x = 6 or x = -2
Check 6, and – 2 are two solutions and satisfy the original
equation
And x-intercepts of the graph are 6, -2
By calculator, draw the graph
Solving Quadratic Equation by factoring
The height h of a baseball t seconds after being hit is given by
h = - 16 t 2 + 64t + 4. When will the baseball reach a height of
64 feet ?
64 = - 16 t 2 + 64t + 4
Standard form 16 t 2 – 64t + 60 = 0
4( 4 t 2 – 16t + 15) = 0 Factor 4 from left side
4(2t – 3)(2t – 5) = 0 Factor the quadratic expression and use
zero factor principle
2t – 3 = 0 or 2t – 5 = 0
Solve each equation
t = 3/2 or t = 5/2 72
64
h = - 16 t 2 + 64t + 4
48
24
0 .5
1
1.5
2 2.5
3
4
Pg 163
Y1 = x2 – 4x + 3
Y2 = 4(x2 – 4x + 3)
Enter window Xmin = -2, Xmax = 8
Ymin = -5 Ymax = 10
And enter graph
Quadratic Equations whose solutions are
given
Example 31, Page 166
Solutions are – 3 and ½, The equation should be in standard
form with integer coefficients
[ x – (-3)] (x – ½) = 0
(x + 3)(x – ½) = 0
x2 – ½ x + 3x – 3/2 = 0
x2 +5x – 3= 0
2
2
2(x2 + 5x –3) = 2(0)
2 2
2 x2 + 5x – 3 = 0
3.3 ,No 48, Page 169
2
I = kCx – k x
2
2
= 0.0002 (6000)x – 0.0002x = 1.2x – 0.0002x
Larger Increase
x – intercept is 6000
i.e neither decrease nor increase
x
0
500
100
0
150
0
200
0
250
0
300
0
350
0
400
0
450
0
500
0
550
0
600
0
650
0
700
0
I
0
550
100
0
135
0
160
0
175
0
180
0
175
0
160
0
135
0
100
0
550
0
-650
140
0
Population 2000 will increase by 1600
1800
Population 7000 will decrease by 1400
1750
1600
1350
1000
500
0
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
3.4 Graphing Parabolas Special cases
• The graph of a quadratic equation is called a parabola
Vertex
y-intercept
x-intercept
x-intercept
x-intercept
x-intercept
y-intercept
Axis of symmetry
Axis of symmetry
Using Graphing Calculator (Page 171)
Enter Y
Y = x2
Y = 3 x2
Y = 0.1 x2
Enter equation
Graph
Enter Graph
Example 3, Pg 177, Finding the vertex of the graph of
y = -1.8x2– 16.2x
Find the x-intercepts of the graph
a) The x-coordinate of the vertex is xv = -b/2a= -(-16.2)/2(-1.8)
• To find the y-coordinate of the vertex, evaluate y at x = -4.5
• yv = -1.8(-4.5)2 – 16.2(-4.5) = 36.45
• The vertex is (-4.5, 36.45)
• b) To find the x-intercepts of the graph, set y = 0 and solve
• - 1.8 x2 – 16.2x = 0
Factor
• -x(1.8x + 16.2) = 0
Set each factor equal to zero
• - x = 0 1.8x + 16.2 = 0
Solve the equation
• x=0
x = -9
• The x-intercepts of the graph are (0,0) and (-9,0)
36
24
12
- 10
-5
0
2
3.6 Quadratic Formula
The solutions of the equation ax 2 + bx + c = 0 with
a = 0 are given by
-b+
b2 – 4ac
2a
•
Complex Numbers
•
For a > 0,  a =
i2 = - 1 or i =
1  a
Discriminant D = b2
1
= i a
- 4ac
If D > 0, the equation has two unequal real solutions
If D = 0, the equation has one real solution of multiplicity two
If D < 0, the equation has two complex (conjugate) solutions
3.6, Page 195, No. 15
Let w represent the width of a pen and l the length of the
enclosure in feet
l
w
Then the amount of chain link fence is given by 4w + 2l = 100
b) 4w +2l = 100
2l = 100 – 4w
l = 50 – 2w
c) The area enclosed is A = wl = w(50 – 2w) = 50w –2w2
The area is 250 feet, so
50w – 2w2 = 250
0 = w 2– 25w + 125
Thus a = 1, b = -25 and c = 125
W = -(-25) +
(- 25) 2 - 4(1)(125)
2(1)
The solutions are 18.09, 6.91 feet
d) l =50 – 2(18.09) = 13.82 feet and l = 50 – 2(6.91) = 36.18 feet
The length of each pen is one third the length of the whole enclosure,
so dimensions of each pen are 18.09 feet by 4.61 feet or 6.91 feet by 12.06 feet
Ex 16, Pg 195
The area of the half circle = 1

1/8 x

2
(1/2 x )2
= 1/2
r=½x
=
r2
2
The area of the rectangle = x2 – 2x
h=x-2
=1
x
2
+
x2 - 2x
8
Total area = 120 square feet
120 = = 1
x
2
x
+ x2 - 2x
8
8(120) = 8 ( 1
x2 + x2 - 2x )
8
0=
x2 + 8 x2 - 16x – 960 , 0 = ( + 8) x2 - 16x – 960
0 = 11.142 x2 – 16x - 960
use quadratic formula , x = 10.03 ft , h = 10.03 – 2 = 8.03ft The overall height of
the window is h + r = h + ½ x = 8.03 + ½ (10.03) = 13.05 ft


